A gas mixture composed of helium and argon has a density of 0.846g/L at a 734mmHg and 298 K.
What is the composition of the mixture by volume?
The general gas law is PV = nRT. n = grams/molar mass which makes this come out to be PV = gRT/M and this can be rearranged to P = gRT/MV where M is molar mass or P*M = gRT. Substitute and solve for M = molar mass. I get about 21 but you need a more accurate number than that estimate.
The let X = fraction He atomic mass 4.003
and 1-X = fraction Ar = atomic mass 39.95
X(4.003) + (1-X)*39.95 = approx 21
Solve for X and 1-X to fractions and multiply by 100 to convert to %.
Why did the helium and argon go into a bar together? Because they wanted to mix it up! *ba dum tss*
But in all seriousness, to find the composition of the mixture by volume, we need to use the ideal gas law equation, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we're given the density (d), we can use the formula: d = (molecular weight * P)/(R * T), where the molecular weight is the average of the molecular weights of helium (4 g/mol) and argon (40 g/mol).
Using the given values, we have:
0.846 g/L = ((0.5 * 4) + (0.5 * 40)) * (734 mmHg / (0.0821 L·atm/(mol·K) * 298 K)
Now we solve for the ratio of helium to argon:
0.846 g/L = (2 + 20) * (734/24.46)
0.846 = 22 * 29.997
0.846 = 659.934
Well, that doesn't seem right. It looks like I took a wrong turn somewhere. Let me recalculate that for you.
To determine the composition of the gas mixture by volume, we can use Dalton's Law of Partial Pressures.
Dalton's Law states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas. The partial pressure of a gas is directly proportional to its mole fraction in the mixture.
Given that the total pressure is 734 mmHg and the mixture is composed of helium and argon, we can assign variables:
Let x be the mole fraction of helium (He).
Let y be the mole fraction of argon (Ar).
According to Dalton's Law, we have:
Partial pressure of helium (P_He) = x * Total pressure
Partial pressure of argon (P_Ar) = y * Total pressure
We also know that the density of the gas mixture is 0.846 g/L. The molar mass of helium is 4.00 g/mol, and the molar mass of argon is 39.95 g/mol.
To find the mole fraction of each gas, we can use the ideal gas law:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Universal gas constant
T = Temperature
Rearranging the equation to solve for n, we have:
n = PV / RT
Given the density (d), we can substitute it into the equation:
n = (m / M) * V, where m is the mass and M is the molar mass of the gas.
For helium:
n_He = (m_He / M_He) * V
For argon:
n_Ar = (m_Ar / M_Ar) * V
Since we have the same Volume, let's compare the number of moles of helium and argon:
n_He = (m_He / M_He) * V = (d * V / M_He) * V = (d * V^2) / M_He
n_Ar = (m_Ar / M_Ar) * V = (d * V / M_Ar) * V = (d * V^2) / M_Ar
Now, using the mole fractions that we defined earlier:
x = n_He / (n_He + n_Ar) and y = n_Ar / (n_He + n_Ar)
Substituting the expressions for n_He and n_Ar, we have:
x = [(d * V^2) / M_He] / [(d * V^2) / M_He + (d * V^2) / M_Ar]
= [1 / M_He] / [1 / M_He + 1 / M_Ar]
= M_Ar / (M_He + M_Ar)
Since the mole fractions of a mixture add up to 1, we can determine the mole fraction of argon:
y = 1 - x
= 1 - (M_Ar / (M_He + M_Ar))
Finally, substituting the known values for the molar masses, we can find the composition of the mixture:
x = M_Ar / (M_He + M_Ar)
= 39.95 g/mol / (4.00 g/mol + 39.95 g/mol)
≈ 0.909
y = 1 - x
= 1 - 0.909
≈ 0.091
Therefore, the composition of the gas mixture by volume is approximately 90.9% helium and 9.1% argon.
To find the composition of the gas mixture by volume, we need to determine the volume fraction of each gas component in the mixture.
First, let's calculate the molar mass of helium (He) and argon (Ar):
- Helium (He):
The molar mass of helium is approximately 4.0026 g/mol.
- Argon (Ar):
The molar mass of argon is approximately 39.948 g/mol.
Next, let's use the ideal gas equation to calculate the number of moles (n) of helium and argon in the mixture:
PV = nRT
Where:
P = Pressure (734 mmHg)
V = Volume (1 L)
n = Number of moles of the gas mixture
R = Gas constant (0.0821 L·atm/(mol·K))
T = Temperature (298 K)
Using the given values, we can rearrange the ideal gas equation to solve for n:
n = PV / RT
n = (734 mmHg * 1 L) / (0.0821 L·atm/(mol·K) * 298 K)
n ≈ 31.17 mol
Now, let's assume the mixture contains x moles of helium and (31.17 - x) moles of argon.
To find the volume fraction of each gas, we'll use the ideal gas equation again:
V = n gas × (RT / P)
The volume fraction (V% ) of each gas component is given by:
V% = (V gas / V total) × 100
Since the total volume is 1 L, we can calculate the volume fraction of each gas as follows:
- Volume fraction of helium (VHe):
VHe = (n He × RT) / P
= (x × 0.0821 L·atm/(mol·K) × 298 K) / 734 mmHg
- Volume fraction of argon (VAr):
VAr = (n Ar × RT) / P
= ((31.17 - x) × 0.0821 L·atm/(mol·K) × 298 K) / 734 mmHg
Finally, we can calculate the composition of the mixture by volume:
Composition of the mixture by volume:
VHe/Vtotal = VHe / (VHe + VAr)
VAr/Vtotal = VAr / (VHe + VAr)
Substituting the values obtained above for VHe and VAr, we can find the numerical composition of the mixture by volume.